In W... alpha
Assuming[x>-0.99,x<0.99];Limit[x^(n+1)/(1-x),n->+Infinity]
remains unevaluated, so Maxima, Sage are nol alone
On 16 Kwi, 08:18, achrzesz <achrz...@wp.pl> wrote:
> One can discuss if in limits of f(x,n) as n-->oo
> x may depend on n or not but in the following version:
>
> sage: assume(x>-0.99,x<0.99)
> sage: n=var('n')
> sage: sage: limit(x^(n+1)/(1-x), n=infinity)
> -limit(x^(n + 1), n, +Infinity)/(x - 1)
>
> sage: assume(x>0)
> sage: sage: limit(x^(n+1)/(1-x), n=infinity)
> 0
>
> the dependence of x on n is not essential
> The limit does not depend on the sign of x
> so the Maxima behaviour inherited by Sage is inconsistent.
>
> On 16 Kwi, 05:56, ancienthart <joalheag...@gmail.com> wrote:
>
> > (Respectfully) Then why does splitting the range of values for x into
> > positive, zero and negative ranges work?
>
> > Joal Heagney
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